Question

Suppose μ is a measure on a measurable space (X,S) and f : X →

[0,[infinity]] is an S-measurable function. Define ν: S →[0,[infinity]] by ν(A)= χ_A
fdμ for A ∈ S. Prove that ν is a measure on (X,S)

Answer 1

To prove that ν is a measure, we need to show that it satisfies non-negativity, countable additivity, and that ν of the empty set is 0.

Step-by-step explanation:

To prove that ν is a measure on (X,S), we need to show that it satisfies the three properties of a measure: non-negativity, countable additivity, and that ν of the empty set is 0.

1. Non-negativity: ν(A) is the integral of the product of the indicator function χA and f dμ. Since both χA and f are non-negative functions, the integral must also be non-negative.

2. Countable Additivity: Let {An} be a countable collection of pairwise disjoint sets in S. We need to show that ν(⋃n=1∞ An) = ∑n=1∞ ν(An). Using linearity of integration, we have ν(A1 ∪ A2 ∪ ...) = ∫ (χA1 ∪ A2 ∪ ...)fdμ

3. ν(∅) = ∫ χ∅fdμ = ∫ 0 dμ = 0